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Question : 13 of 36

Marks: +1, -0

Solve the following equation using the method of matrix inversion:

2 x-3 y+5=0

3 x+y=9

$x=3, y=2$
$x=4, y=2$
$x=2, y=3$
$x=2, y=4$

Solution:

Concept:

Represent the system as

AX = B

and solve using

X = A^{-1}B

.

Explanation:

Rewrite equations:

2x - 3y = -5

and

3x + y = 9

.

Matrix form:

A = \begin{pmatrix} 2 & -3 \\ 3 & 1 \end{pmatrix},\quad X = \begin{pmatrix} x \\ y \end{pmatrix},\quad B = \begin{pmatrix} -5 \\ 9 \end{pmatrix}

Compute determinant:

\det(A) = 2(1) - (-3)(3) = 2 + 9 = 11

.

Find inverse:

A^{-1} = \frac{1}{11} \begin{pmatrix} 1 & 3 \\ -3 & 2 \end{pmatrix}

.

Multiply:

X = A^{-1}B = \frac{1}{11} \begin{pmatrix} 1 & 3 \\ -3 & 2 \end{pmatrix} \begin{pmatrix} -5 \\ 9 \end{pmatrix} = \frac{1}{11} \begin{pmatrix} 1(-5)+3(9) \\ -3(-5)+2(9) \end{pmatrix} = \frac{1}{11} \begin{pmatrix} -5+27 \\ 15+18 \end{pmatrix} = \frac{1}{11} \begin{pmatrix} 22 \\ 33 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}

.

Thus

x=2

,

y=3

.

Answer:

x=2,\ y=3

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